9.5 The Binomial Expansion

Earlier we studied products of polynomials, and in particular we found expanded forms for powers of binomials such as $(a + b)^{2}$ and $(a + b)^{3}$ . The amount of work involved in expanding such powers increases as the exponent gets larger. In this section we will learn how to raise a binomial to any positive integer power, without having to perform the polynomial products.

Investigation

Powers of Binomials

In this investigation we will look for patterns in the expansion of $(a + b)^{n}$ . We begin by computing several such powers.

Expand each power and fill in the blanks. Arrange the terms in each expansion in descending powers of $a$ .

$(a + b)^{0} =$ ________________________
$(a + b)^{1} =$ ________________________
$(a + b)^{2} =$ ________________________
$(a + b)^{3} =$ ________________________
Hint: Start by writing
$(a + b)^{3} = (a + b) (a + b)^{2}$
and use your answer to #3.
$(a + b)^{4} =$ ________________________
Hint: Start by writing
$(a + b)^{4} = (a + b) (a + b)^{4}$
and use your answer to #4.
$(a + b)^{5} =$ ________________________

Do you see a relationship between the exponent

n

and the number of terms in the expansion of

(a + b)^{n}

? (Notice that for

n equals 0

we have

(a + b)^{0} = 1

, which has one term.) Fill in the table below.

$n$	Number of terms in $(a + b)^{n}$
$0$
$1$
$2$
$3$
$4$
$5$

First observation: In general, the expansion of $(a + b)^{n}$ has ________________________ terms.

Next we'll consider the exponents on

a

and

b

in each term of the expansions. Refer to your expanded powers in parts 1-5, and fill in the next table.

$n$	First term of $(a + b)^{n}$	Last term of $(a + b)^{n}$	Sum of exponents on $a$ and $b$ in each term
$0$
$1$
$2$
$3$
$4$
$5$

Second observation: In any term of the expansion of $(a + b)^{n}$ , the sum of the exponents on $a$ and on $b$ is ________________________

In fact, we can be more specific in describing the exponents in the expansions. We will use

k

to label the terms in the expansion of

(a + b)^{n}

, starting with

k equals 0

. For example, for

n equals 2

we label the terms as follows: .

$\begin{aligned} (a + b)^{2} = & a^{2} + 2 a b + b^{2} \\ k = 0 k = 1 k = 2 \end{aligned}$

We can make a table showing the exponents on

a

and

b

in each term of

(a + b)^{2}

:
Case

n equals 2

$k$	$0$	$1$	$2$
Exponent on $a$	$2$	$1$	$0$
Exponent on $b$	$0$	$1$	$2$

Complete the tables shown below for the cases

n = 3, n = 4

and

n equals 5

.
Case

n equals 3

$k$	$0$	$1$	$2$	$3$
Exponent on $a$	$3$
Exponent on $b$	$0$

Case

n equals 4

$k$	$0$	$1$	$2$	$3$	$4$
Exponent on $a$	$4$
Exponent on $b$	$0$

Case

n equals 5

$k$	$0$	$1$	$2$	$3$	$4$	$5$
Exponent on $a$	$4$
Exponent on $b$	$0$

Third observation: The variable factors of the $k^{th}$ term in the expansion of $(a + b)^{n}$ may be expressed as ________________________. (Fill in the correct powers in terms of $k$ and $n$ for $a$ and $b$ .)

In the next investigation we will look for patterns in the coefficients of the terms of the expansions.

Powers of Other Binomials

We can use what we learned from the Investigation to raise other binomials to powers.

Example 1

Expand each power of a binomial.

$open parenthesis x plus 1 close parenthesis cubed$
$(2 m - n)^{4}$

We know from the Investigation that
$\begin{matrix} (a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} \end{matrix}$
We replace $a$ with $x$ and $b$ with 1 to get
$\begin{aligned} (x + 1)^{3} & = x^{3} + 3 x^{2} (1) + 3 x (1)^{2} + 1^{3} \\ = x^{3} + 3 x^{2} + 3 x + 1 \end{aligned}$
This time we take the expansion
$\begin{matrix} (a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} \end{matrix}$
from the Investigation and let $a equals 2 m$ and $b = - n$ .
$\begin{aligned} (2 m - n)^{4} & = (2 m)^{4} + 4 (2 m)^{3} (- n) + 6 (2 m)^{2} (- n)^{2} + \\ + 4 (2 m) (- n)^{3} + (- n)^{4} \\ = 16 m^{4} + 4 (8 m^{3}) (- n) + 6 (4 m^{2}) (n^{2}) + \\ + 4 (2 m) (- n^{3}) + n^{4} \\ = 16 m^{4} - 32 m^{3} n + 24 m^{2} n^{2} - 8 m n^{3} + n^{4} \end{aligned}$

Expand $(r + 2 s)^{3} =$ _____

$r^{3} + 6 r^{2} s + 12 r s^{2} + 8 s^{3}$

Expand $open parenthesis r plus 2 s close parenthesis cubed$

$r^{3} + 6 r^{2} s + 12 r s^{2} + 8 s^{3}$

The Binomial Coefficient

Now we continue our study of powers of binomials. What should the expansion of $(a + b)^{6}$ look like? From the Iinvestigation we know that it should have 7 terms, with the exponents on $a$ decreasing from 6 down to 0, and the exponents on $b$ increasing from 0 up to 6.

Thus, the expansion should have the form

$\begin{aligned} (a + b)^{6} & = \underset{―}{} a^{6} + \underset{―}{} a^{5} b + \underset{―}{} a^{4} b^{2} + \underset{―}{} a^{3} b^{3} + \\ + \underset{―}{} a^{2} b^{4} + \underset{―}{} a b^{5} + \underset{―}{} b^{6} \end{aligned}$

But what about the numerical coefficients of the terms? We will use the notation $_{n} C_{k}$ for the coefficient of the $k^{th}$ term in the expansion of $(a + b)^{n}$ . For example, in the expansion of $(a + b)^{6}$ above,

the coefficient of $a to the power 6$ is denoted by $sub 6 C sub 0$ ,
the coefficient of $a^{5} b^{1}$ is $sub 6 C sub 1$ ,
the coefficient of $a to the power 4 b squared$ is $sub 6 C sub 2$ ,

and so on. Note that the 6 in front of the $C$ indicates that $n equals 6$ , and the number following $C$ corresponds to the exponent on $b$ . The symbol $_{n} C_{k}$ is called the binomial coefficient.

Example 2

Use your expansions from the Investigation to evaluate the following binomial coefficients.

$sub 4 C sub 3$
$sub 5 C sub 2$

$sub 4 C sub 3$ is the coefficient of the term containing $b^{3}$ in the expansion of $(a + b)^{4}$ . Referring to Step 5 in the Investigation, we see that the coefficient of the term $4 a b cubed$ is 4. Thus, $sub 4 C sub 3 equals 4$ .
$sub 5 C sub 2$ is the coefficient of the term containing $b^{2}$ in the expansion of $(a + b)^{5}$ . Referring to Step 6 in the Investigation, we see that the coefficient of the term $10 a cubed b squared$ is 10. Thus, $sub 5 C sub 2 equals 10$ .

Evaluate the binomial coefficients.

$sub 5 C sub 4 equals$ _____
$sub 5 C sub 0 equals$ _____

Evaluate the binomial coefficients.

$sub 5 C sub 4$
$sub 5 C sub 0$

Investigation

Pascal's Triangle

To get a clearer picture of the binomial coefficients, consider again the expansions of $(a + b)^{n}$ you calculated in the last Investigation, but this time look only at the numerical coefficients of each term:

$n equals 0$		$1$
$n equals 1$		$1 1$
$n equals 2$		$1 2 1$
$n equals 3$		$1 3 3 1$
$n equals 4$		$1 4 6 4 1$
$n equals 5$		$1 5 10 10 5 1$

This triangular array of numbers is known as Pascal's triangle. It has many interesting and surprising properties that have been extensively studied. We might first make the following observations.

Each row of Pascal's triangle begins with the number ______ and ends with the number ______.
The second number and the next-to-last number in the $n^{th}$ row are _____.
There is an interesting pattern for the rest of the numbers in each row. Pick any number in the row $n equals 4$ , and look at the two closest numbers in the previous row. (For example, if you picked 6, the two closest numbers in the previous row are 3 and 3.) Do you see a relationship between the numbers? Try the same thing with several numbers in row $n equals 5$ to test your theory.
Starting with the row $n equals 2$ , any number in the triangle (except the first and last 1's in each row) can be found by ____________________.
Using your answer to question #3, continue Pascal's triangle to include the row for $n equals 6$ .
The numbers in Pascal's triangle are the binomial coefficients we have been looking for. Specifically, the number in the $k^{th}$ position (starting with $k equals 0$ ) of the $n^{th}$ row of the triangle is $_{n} C_{k}$ . Thus, we can use the numbers in Pascal's triangle to expand $(a + b)^{n}$ .
Use Pascal's triangle to find the binomial coefficient $sub 6 C sub 4$ .
Expand: $(a + b)^{6}$ : ________________________
Expand: $(x - 2)^{6}$ : ________________________. (Hint: Use your answer to #6, replacing $a$ by $x$ and $b$ by $minus 2$ .)
Continue Pascal's triangle to include the row for $n equals 7$ .

Using Pascal's Triangle

We can now write the expanded form of any binomial power without having to perform the multiplication.

Example 3

Find the expanded form of $(3 r - q)^{6}$ without performing the multiplication.

$\begin{aligned} (a + b)^{6} & = a^{6} + \underset{―}{} a^{5} b + \underset{―}{} a^{4} b^{2} + \underset{―}{} a^{3} b^{3} + \\ + \underset{―}{} a^{2} b^{4} + \underset{―}{} a b^{5} + b^{6} \end{aligned}$

Next, fill in the blanks with the binomial coefficients from Pascal's triangle.

$\begin{matrix} (a + b)^{6} = a^{6} + 6 a^{5} b + 15 a^{4} b^{2} + 20 a^{3} b^{3} + 15 a^{2} b^{4} + 6 a b^{5} + b^{6} \end{matrix}$

Finally, replace $a$ by $3 r$ and replace $b$ by $- q$ , then simplify to find

$\begin{aligned} (3 r - q)^{6} & = (3 r)^{6} + 6 (3 r)^{5} (- q) + 15 (3 r)^{4} (- q)^{2} + 20 (3 r)^{3} (- q)^{3} + \\ + 15 (3 r)^{2} (- 6)^{4} + 6 (3 r) (- q)^{5} + (- q)^{6} \\ = 729 r^{6} - 1458 r^{5} q + 1215 r^{4} q^{2} - 540 r^{3} q^{3} + \\ + 135 r^{2} q^{2} - 18 r q^{5} + q^{6} \end{aligned}$

Expand $(w + 10 z)^{6}$ .

_____

Do not enter any commas. For example, use "10000" rather than "10,000".

$\begin{aligned} w^{6} + 60 w^{5} z + 1500 w^{4} z^{2} + 20, 000 w^{3} z^{3} & + 150, 000 w^{2} z^{4} + \\ + 600, 000 w z^{5} + 1, 000, 000 z^{6} \end{aligned}$

Expand $(w + 10 z)^{6}$ .

$\begin{aligned} w^{6} + 60 w^{5} z + 1500 w^{4} z^{2} + 20, 000 w^{3} z^{3} & + 150, 000 w^{2} z^{4} + \\ + 600, 000 w z^{5} + 1, 000, 000 z^{6} \end{aligned}$

Factorial Notation

Finding binomial coefficients from Pascal's triangle is tedious for large values of $n$ . It would be convenient to have a formula for calculating the coefficients directly. For this formula we need a new symbol, $n!$ (read " $n$ factorial"), which is defined as follows:

For example,

$6! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720 and 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

Evalute $4 \cdot 3! =$ _____

Evalute $4 times 3 !$

Write the first three factors of

$2 n !$
_____
$open parenthesis 2 n close parenthesis !$
_____

$2 \cdot n \cdot (n - 1)$
$(2 n) \cdot (2 n - 1) \cdot (2 n - 2)$

Write the first three factors of

$2 n !$
$open parenthesis 2 n close parenthesis !$

$2 \cdot n \cdot (n - 1)$
$(2 n) \cdot (2 n - 1) \cdot (2 n - 2)$

Note that $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ can be written as

$7 \cdot 6! = 7 \cdot (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)$

and $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ can be written as

$5 \cdot 4! = 5 \cdot (4 \cdot 3 \cdot 2 \cdot 1)$

In general, we have the following relationship:

This relationship can be helpful in simplifying expressions involving factorials.

Example 6

Write each expression in expanded form and simplify.

$\frac{6!}{3!}$
$\frac{4! 6!}{8!}$

$\frac{6!}{3!} = \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3!} = 6 \cdot 5 \cdot 4 = 120$
$\frac{4! 6!}{8!} = \frac{4 \cdot 3 \cdot 2 \cdot 1 \cdot 6!}{8 \cdot 7 \cdot 6!} = \frac{3}{7}$

Write $\frac{5!}{3! 2!}$ in expanded form and simplify.

Simplified form: _____

$10$

Write $\frac{5!}{3! 2!}$ in expanded form and simplify.

$10$

We have defined $n!$ when $n$ is a positive integer. What if $n equals 0$ ? To be consistent with the equation $n! = n \cdot (n - 1)!$ we must have

$1! = 1 \cdot (1 - 1)! or 1! = 1 \cdot (0)!$

This leads us to the following definition.

Note that both $1 !$ and $0 !$ are equal to $1$ .

The Binomial Coefficient in Factorial Notation

Earlier we introduced the notation $_{n} C_{k}$ for the numerical coefficients in the expansion of $(a + b)^{n}$ . These binomial coefficients, which are given by Pascal's triangle, can also be expressed using factorial notation as follows.

Example 7

Evaluate each binomial coefficient.

$sub 6 C sub 2$
$sub 9 C sub 8$

$_{6} C_{2} = \frac{6!}{(6 - 2)! 2!} = \frac{6!}{4! 2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15$
$_{9} C_{8} = \frac{9!}{(9 - 8)! 8!} = \frac{9!}{1! 8!} = \frac{9 \cdot 8!}{1 \cdot 8!} = 9$

Evaluate the binomial coefficient $sub 8 C sub 6 equals$ _____

$10$

Evaluate the binomial coefficient $sub 8 C sub 6$ .

$10$

We now have two methods for computing the binomial coefficient $_{n} C_{k}$ : Pascal's triangle and our formula using factorials. When $n$ is small, especially if all the coefficients of the binomial expansion are needed, Pascal's triangle is often the easier choice. But when $n$ is large or if only one coefficient is required, the factorial formula for $_{n} C_{k}$ is probably quicker than Pascal's triangle.

Example 8

Find the coefficient of $m to the power 11 n cubed$ in the expansion of $(m + n)^{14}$ .

We are simply replacing $a$ with $m$ and $b$ with $n$ in the expansion of $(a + b)^{14}$ , so the term containing $m to the power 11 n cubed$ has $n equals 14$ and $k equals 3$ , and its coefficient is $sub 14 C sub 3$ .

$_{14} C_{3} = \frac{14!}{(14 - 3)! 3!} = \frac{14 \cdot 13 \cdot 12 \cdot 11!}{11! 3!} = \frac{14 \cdot 13 \cdot 12}{3 \cdot 2 \cdot 1} = 364$

The coefficient of $m to the power 11 n cubed$ is $364$ .

Find the coefficient of $s cubed t to the power 4$ in the expansion of $(s + t)^{7}$ .

Answer: _____

$35$

Find the coefficient of $s cubed t to the power 4$ in the expansion of $(s + t)^{7}$ .

$35$

The Binomial Theorem

Consider the expansion of $(a + b)^{6}$ written with the $_{n} C_{k}$ notation for the coefficients.

$\begin{aligned} (a + b)^{6} & = a^{6} b^{0} +_{6} C_{1} a^{5} b^{1} +_{6} C_{2} a^{4} b^{2} +_{6} C_{3} a^{3} b^{3} +_{6} C_{4} a^{2} b^{4} +_{6} C_{5} a^{1} b^{5} +_{6} C_{6} a^{0} b^{6} \\ k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 k = 6 \end{aligned}$

Check that each term can be written in the form

$\begin{matrix} _{6} C_{k} a^{6 - k} b^{k} \end{matrix}$

for $k equals 0$ to $k equals 6$ . This means that we can use sigma notation to write the sum of the terms in the expansion.

$\begin{matrix} (a + b)^{6} = \sum_{k = 0}^{6}_{6} C_{k} a^{6 - k} b^{k} \end{matrix}$

Notice that the sigma notation indicates terms for $k equals 0$ to $k equals 6$ , so there are 7 terms in the expansion, as there should be. Also note that the sum of the exponents on $a$ and $b$ is $(6 - k) + k = 6$ .

The equation above is a special case of the binomial theorem, which uses everything we have learned to write the expanded form of a power of a binomial in the most compact form possible. We can write the general form of the theorem, for positive integers $n$ , as follows.

Example 9

Use sigma notation to write the expanded form for $(x - 2 y)^{10}$ .
Find the term containing $y to the power 7$ , and simplify.

We apply the binomial theorem with $n equals 10$ , replacing $a$ by $x$ and $b$ by $minus 2 y$ , to get
$(x - 2 y)^{10} = \sum_{k = 0}^{10}_{10} C_{k} x^{10 - k} (- 2 y)^{k}$
The term containing $y to the power 7$ corresponds to $k equals 7$ . The $k equals 7$ term looks like
$_{10} C_{7} x^{10 - 7} (- 2 y)^{7} or_{10} C_{7} x^{3} (- 2 y)^{7}$
The value of $sub 10 C sub 7$ is
$_{10} C_{7} = \frac{10!}{(10 - 7)! 7!} = \frac{10 \cdot 9 \cdot 8 \cdot 7!}{3! 7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$
and $(- 2 y)^{7} = - 128 y^{7}$ . Thus, the term we want is
$_{10} C_{7} x^{3} (- 2 y)^{7} = 120 x^{3} (- 128 y^{7}) = - 15, 360 x^{3} y^{7}$

Use sigma notation to write the expanded form for $(2 m - 3 n)^{7}$ .
_____
Find the term containing $m^{3}$ , and simplify.
_____

$(2 m - 3 n)^{7} = \sum_{k = 0}^{7}_{7} C_{k} (2 m)^{7 - k} (- 3 n)^{k}$
$_{7} C_{4} (2 m)^{3} (- 3 n)^{4} = 22, 680 m^{3} n^{4}$

Use sigma notation to write the expanded form for $(2 m - 3 n)^{7}$ .
Find the term containing $m^{3}$ , and simplify.

$(2 m - 3 n)^{7} = \sum_{k = 0}^{7}_{7} C_{k} (2 m)^{7 - k} (- 3 n)^{k}$
$_{7} C_{4} (2 m)^{3} (- 3 n)^{4} = 22, 680 m^{3} n^{4}$

Section Summary

Vocabulary

Binomial coefficient
Pascal's triangle
Factorial

CONCEPTS

The binomial coefficient $_{n} C_{k}$ is the coefficient of the term containing $b^{k}$ in the expansion of $(a + b)^{n}$ .
The numbers in Pascal's triangle are the binomial coefficients. Specifically, the number in the $k^{th}$ position (starting with $k equals 0$ ) of the $n^{th}$ row of the triangle is $_{n} C_{k}$ .
If $n$ is a positive integer,
$n! = n \cdot (n - 1) \cdot (n - 2) \cdot \dots \cdot 3 \cdot 2 \cdot 1$
For $n greater than or equal to 0$ and $0 less than or equal to k less than or equal to n$ ,
$_{n} C_{k} = \frac{n!}{(n - k)! k!}$
The Binomial Theorem
$(a + b)^{n} = \sum_{k = 0}^{n}_{n} C_{k} a^{n - k} b^{k}$

STUDY QUESTIONS

What does the symbol $_{n} C_{k}$ denote?
Explain how to use Pascal's triangle when expanding a power of a binomial.
What is factorial notation?
State a formula for the binomial coefficients in factorial notation.
Explain how to use the binomial theorem to expand $(a + b)^{7}$ .

SKILLS

Practice each skill in the Homework Problems listed.

Expand powers of binomials: #1–22, 33–34
Simplify factorial notation: #23–24
Evaluate binomial coefficients: #25–26
Find specific terms of binomial expansions: #28–32, #35-42
Use Pascal's triangle: #43–44

Homework 9.5

For Problems 1–4, answer the questions without expanding the power.

How many terms are in the expansion of $(a + b)^{50}$ ? In the expansion of $(2 x + 3 y)^{100}$ ?

$51; 101$

How many terms are in the expansion of $(a + b)^{75}$ ? In the expansion of $(5 x - 7 y)^{200}$ ?

What is the sum of the exponents on $x$ and $y$ in each term of the expansion of $(x + y)^{100}$ ? In the expansion of $(8 x - 7 y)^{50}$ ?

$100; 50$

What is the sum of the exponents on $x$ and $y$ in each term of the expansion of $(x + y)^{200}$ ? In the expansion of $(9 x + 4 y)^{75}$ ?

Write down the portion of Pascal's triangle corresponding to rows from $n equals 0$ to $n equals 10$ . How many rows are involved?

11 rows

$n equals 0$		$1$
$n equals 1$		$1 1$
$n equals 2$		$1 2 1$
$n equals 3$		$1 3 3 1$
$n equals 4$		$1 4 6 4 1$
$n equals 5$		$1 5 10 10 5 1$
$n equals 6$		$1 6 15 20 15 6 1$
$n equals 7$		$1 7 21 35 35 21 7 1$
$n equals 8$		$1 8 28 56 70 56 28 8 1$
$n equals 9$		$1 9 36 84 126 126 84 36 9 1$
$n equals 10$		$1 10 45 120 210 252 210 120 45 10 1$

Write down the portion of Pascal's triangle corresponding to rows from $n equals 0$ to $n equals 12$ . How many rows are involved?

For Problems 7–18, write the power in expanded form.

$(x + 3)^{5}$

$x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 243$

$(2 x + y)^{4}$

$(z - 3)^{4}$

$z^{4} - 12 z^{3} + 54 z^{2} - 108 z + 81$

$(2 w - 1)^{5}$

${(2 x - \frac{y}{2})}^{3}$

$8 x^{3} - 6 x^{2} y + \frac{3}{2} x y^{2} - \frac{1}{8} y^{3}$

${(\frac{x}{3} + 3)}^{6}$

$(x^{2} - 3)^{7}$

$x^{14} - 21 x^{12} + 189 x^{10} - 945 x^{8} + 2835 x^{6} - 5103 x^{4} + 5103 x^{2} - 2187$

${(1 - y^{2})}^{5}$

$(x + y)^{5}$

$x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5}$

$(x - y)^{6}$

$(p - 2 q)^{4}$

$p^{4} - 8 p^{3} q + 24 p^{2} q^{2} - 32 p q^{3} - 16 p^{4}$

$(m + 3 n)^{8}$

Simplify $(1 + 5 t)^{3} + (1 - 5 t)^{3}$ .

$2 plus 150 t squared$

Simplify $(3 + 2 a)^{4} + (3 - 2 a)^{4} .$

Write ${(z - \frac{1}{z})}^{5}$ in descending powers of $z$ .

$z^{5} - 5 z^{3} + 10 z - 10 z^{- 1} + 5 z^{- 3} - z^{- 5}$

Write ${(v + \frac{1}{v})}^{4}$ in descending powers of $v$ .

Write the expressions in Problems 23–24 in expanded form and simplify.

$5 !$
$\frac{9!}{7!}$
$\frac{5! 7!}{12!}$
$\frac{8!}{2! (8 - 2)!}$

$5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$
$\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 72$
$\frac{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1}{792}$
$\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1) (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)} = 28$

$7 !$
$\frac{12!}{11!}$
$\frac{12! 4!}{16!}$
$\frac{10!}{4! (10 - 4)!}$

In Problems 25–26, evaluate the binomial coefficient.

$sub 9 C sub 6$
$sub 12 C sub 3$
$sub 20 C sub 18$
$sub 14 C sub 9$

$84$
$220$
$190$
$2002$

$sub 8 C sub 5$
$sub 13 C sub 4$
$sub 18 C sub 16$
$sub 16 C sub 7$

Find in ascending powers of $x$ the first three terms in the expansions of

$(1 - 2 x)^{7}$
$(2 - x)^{6}$

$1 - 14 x + 84 x^{2}$
$64 - 192 x + 240 x^{2}$

Find in descending powers of $y$ the first three terms in the expansions of

$(5 y - 2)^{4}$
$(3 y + 1)^{5}$

Find in descending powers of $u$ the first three non-zero terms of $(4 u - 1)^{5} - (4 u + 1)^{5}$ .

$- 2560 u^{4} - 320 u^{2} - 2$

Find in ascending powers of $t$ the first three non-zero terms of $(1 - 3 t)^{6} - (1 + 3 t)^{6}$ .

Find the first four terms of

$(1 - 5 c)^{6}$
$(1 - 4 c) (1 - 5 c)^{6}$

$1 - 30 c + 375 c^{2} - 2500 c^{3}$
$1 - 34 c + 495 c^{2} - 4000 c^{3}$

Find the first four terms of

$(1 + 2 a)^{9}$
$(1 + 3 a) (1 + 2 a)^{9}$

Simplify ${(1 + \sqrt{3})}^{4} + {(1 - \sqrt{3})}^{4}$ .

$56$

Simplify ${(\sqrt{2} + \sqrt{3})}^{4} + {(\sqrt{2} - \sqrt{3})}^{4}$ .

For Problems 35–42, find the coefficient of the indicated term.

$(x + y)^{20}; x^{13} y^{7}$

$77, 520$

$(x + y)^{15}; x^{12} y^{3}$

$(a - 2 b)^{12}; a^{5} b^{7}$

$minus 101, 376$

$(2 a - b)^{12}; a^{8} b^{4}$

${(x - \sqrt{2})}^{10}; x^{4}$

$1680$

${(x - \frac{1}{2})}^{8}; x^{5}$

${(a^{3} - b^{3})}^{9}; a^{18} b^{9}$

$minus 84$

${(x^{2} - y^{2})}^{7}; x^{10} y^{4}$

For Problems 43–44, refer to Pascal's triangle to answer the questions.

Write out the terms of the sequence $11 to the power n$ for $n equals 0, 1, 2, 3, 4$ . How are these related to the rows of Pascal's triangle? Can you explain why? (Hint: Rewrite $11$ as $10 plus 1$ .)

$1, 11, 121, 1331, 14641.$ The digits of the terms in the sequence correspond to the numbers in the first five rows of Pascal's triangle. If $11^{n} = (10 + 1)^{n}$ is expanded as a binomial, each term is the product of a number from Pascal's triangle times a power of 10 times a power of 1.

Write out the terms of the sequence $1.1 to the power n$ for $n equals 0, 1, 2, 3, 4$ . How are these related to the rows of Pascal's triangle? Can you explain why? (Hint: Rewrite $1.1$ as $1 plus 0.1$ .)

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.